First order differential equation.

it is difficult to read the question that way round

Reply 2

user2020user

Original post by Branny101

...

This is a separable first order differential equation of the form:

\begin{aligned} \dfrac{\text{d}v}{\text{d}t} = f(v) \cdot g(t) \implies \displaystyle \int \dfrac{1}{f(v)} \text{ d}v = \int g(t) \text{ d}t \end{aligned}

Spoiler

(edited 9 years ago)

Reply 3

Original post by Khallil

This is a separable first order differential equation of the form:

\begin{aligned} \dfrac{\text{d}v}{\text{d}t} = f(v) \cdot g(t) \implies \displaystyle \int \dfrac{1}{f(v)} \text{ d}v = \int g(t) \text{ d}t \end{aligned}

Does this help?

ahhh !! much better :wink:

Reply 4

user2020user

Original post by the bear

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Fix the spoiler! It's exploded :eek:

Reply 5

[QUOTE="bear;47852732" the="the"]

Original post by Khallil

This is a separable first order differential equation of the form:

\begin{aligned} \dfrac{\text{d}v}{\text{d}t} = f(v) \cdot g(t) \implies \displaystyle \int \dfrac{1}{f(v)} \text{ d}v = \int g(t) \text{ d}t \end{aligned}

Does this help?

ahhh !! much better :wink:

Sorry guys, phone's being tedious... And hmm one second let me just try this

Posted from TSR Mobile

Reply 6

it is Separation of Variables...

you must keep the (10 -5V) together as you move it

(edited 9 years ago)

Reply 7

Original post by Khallil

This is a separable first order differential equation of the form:

\begin{aligned} \dfrac{\text{d}v}{\text{d}t} = f(v) \cdot g(t) \implies \displaystyle \int \dfrac{1}{f(v)} \text{ d}v = \int g(t) \text{ d}t \end{aligned}

Yep, Khallil is right. However, just to make things clearer:

10 - 5v = 2(5 - v)

Divide through by (5 - v) and now you can solve using separation of variables.

(edited 9 years ago)

Reply 8

Original post by the bear

it is Separation of Variables...

you must keep the (10 -5V) together as you move it

Original post by VannR

Yep, Khallil is right. However, just to make things clearer:

10 - 5v = 2(5 - v)

Divide through by (5 - v) and now you can solve using separation of variables.

Ahh and just to verify why can you not add the 5v onto both sides? Other than the fact that it won't be in the form they'd like?

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Reply 9

Original post by Branny101

Ahh and just to verify why can you not add the 5v onto both sides? Other than the fact that it won't be in the form they'd like?

Posted from TSR Mobile

Separation of Variables is always by timesing or dividing, never by adding or subtraction

:mob:

Reply 10

Original post by Branny101

Ahh and just to verify why can you not add the 5v onto both sides? Other than the fact that it won't be in the form they'd like?

Posted from TSR Mobile

Adding 5v to both sides in this case would be pointless, since the equation can be solved using separation of variables.

If you had a situation where you had, say, x.dv/dx in your equation or something similar, you may want to add 5v to both sides to solve as an exact equation. This is not the case here.

Reply 11

Original post by VannR

Original post by the bear

Separation of Variables is always by timesing or dividing, never by adding or subtraction

:mob:

Ahh so my order of thinking should be separating --> then rearranging?

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Reply 12

Original post by Branny101

Ahh so my order of thinking should be separating --> then rearranging?

Posted from TSR Mobile

Yes. The order in which you should try methods is:

1. Separation of variables. It is usually pretty easy to spot if this can be done or not.
2. Exact equation. Check if the equation is exact.
3. Find and use an integrating factor. Remember to express the equation in linear form first.

Just realised that you might not be doing FP2, and this might be C4 differentiation :rolleyes:

. In that case, you'll only ever have to use separation of variables.

Reply 13

ΘTheta

[QUOTE="Branny101;47852741"]

Original post by the bear

Sorry guys, phone's being tedious... And hmm one second let me just try this

Posted from TSR Mobile

Hello There :smile:

For part b, is the answer t=0.277 (3sf)?
and for c, do you draw the graph of v=2(1-e^-5t), with t on the x axis and v on the y axis, and since that there is an asymptope an v=2, hence the speed can never exceed 2ms^-1? Thanks ! :biggrin:

Reply 14

[QUOTE="ΘTheta;47853265"]

Original post by Branny101

Hello There :smile:

The graph isn't given in the answers but that's what I got for both :smile:

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(edited 9 years ago)

Reply 15

Original post by VannR

. In that case, you'll only ever have to use separation of variables.

C4 only dude, so I'll only be separating? You sure? I'm on the AQA spec

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Reply 16

ΘTheta

[QUOTE="Branny101;47854399"]

Original post by ΘTheta

The graph isn't given in the answers but that's what I got for both :smile:

and yes to part b

Posted from TSR Mobile

There is no graph in the answer scheme for the question in the book? :eek:

That odd, the question says to draw a graph???
Did you get the asymptope part of the question c.? :smile:

Reply 17